differential equations

Answer q1.
dy/dx= -y²/x(x-y)
Take y=vx
dy/dx=v+x.dv/dx
v+x.dv/dx=-v²/(v-1)
xdv/dx=v/v-1
dx/x=(v-1)/v dv
integrating on both sides
lnx=v-lnv
substituting v=y/x
lnx=y/x -lny+lnx
lny=y/x ........................(answer)

2nd one is really easy!

take x²+y²=t²
then xdx+ydy=2tdt
and multiply both the sides by x²
we get
2tdt/d(y/x)=√(a²-t²)/(1+(y/x)²)

hence

(2t/√a²-t²)dt=d(y/x)/(1+(y/x)²)

put √a²-t²=z

integrating we have

-2√z=tan^-1(y/x)+c

Nicely Done Rohan... I was doing the same but got stuck in between..

Anyway well done pal...